OCR MEI Further Pure Core 2019 June — Question 15 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeStandard integral of 1/√(x²+a²)
DifficultyChallenging +1.2 This is a Further Maths question requiring completion of the square to transform the integrand into standard form, then applying the inverse hyperbolic function formula (or equivalent logarithmic form). While it involves multiple steps (algebraic manipulation, recognizing the standard integral, and evaluating at limits), these are well-practiced techniques for Further Maths students. The 'show that' format with a specific answer provides guidance, making it moderately above average difficulty but not requiring novel insight.
Spec4.08h Integration: inverse trig/hyperbolic substitutions
15 In this question you must show detailed reasoning. Show that \(\int _ { \frac { 3 } { 4 } } ^ { \frac { 3 } { 2 } } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 4 x + 2 } } \mathrm {~d} x = \frac { 1 } { 2 } \ln \left( \frac { 3 + \sqrt { 5 } } { 2 } \right)\).
Question 15:
AnswerMarks
15DR
3 3
1 1
2 dx2 dx
3 3
4x2 4x2 (2x1)21
AnswerMarks Guidance
4 4M1 3.1a
1 1
or 2 dx
2 3 (x1/2)21/4
4
attempt to complete the
square
3
1 2
 arsinh(2x1)
 
2 3
AnswerMarks
41.1b
1.1barsinh(2x  1) (oe)
 ½ oe e.g. ln form2
1 
  arsinhu  if u =2x1
2 1
2
AnswerMarks
M1arsinh(2x  1) (oe)
A1 ½ oe e.g. ln form
(3)
1 1
 [arsinh(2)arsinh( )]
2 2
1 1 5
 [ln(2 5)ln(  )]
AnswerMarks Guidance
2 2 2M1
A11.1b arsinh x=ln(x+(x2 + 1))
correct expression(used)
1 2( 52)
 ln
AnswerMarks Guidance
2 51M1 2.1
1 2( 52)( 51)
 ln
AnswerMarks Guidance
2 ( 51)( 51)M1 2.1
(must be seen)
1 ( 53)
 ln *
AnswerMarks
2 21 ( 53)
 ln *
AnswerMarks Guidance
2 2A1cao 2.1
(5)
[8]
Alternative solution
x + ay = 2, x + ay + z = 1
 2x + z = 1, z = 2x 1
x + ay = 2  y = (2+ x)/a
3x6
 2x 12x3
a
2a6 2 4a9
x y , z = 
3 3 3
M1
M1
M1
A3
[6]
from 2 equations
to get eqn in one unknown
DR
3 3
1 1
2 dx2 dx
3 3
4x2 4x2 (2x1)21
4 4
M1
3.1a
3
1 1
or 2 dx
2 3 (x1/2)21/4
4
1.1b
1.1b
M1
A1
arsinh x=ln(x+(x2 + 1))
correct expression
rationalizing denominator
(must be seen)
Question 15:
15 | DR
3 3
1 1
2 dx2 dx
3 3
4x2 4x2 (2x1)21
4 4 | M1 | 3.1a | 3
1 1
or 2 dx
2 3 (x1/2)21/4
4
attempt to complete the
square
3
1 2
 arsinh(2x1)
 
2 3
4 | 1.1b
1.1b | arsinh(2x  1) (oe)
 ½ oe e.g. ln form | 2
1 
  arsinhu  if u =2x1
2 1
2
M1 | arsinh(2x  1) (oe)
A1 |  ½ oe e.g. ln form
(3)
1 1
 [arsinh(2)arsinh( )]
2 2
1 1 5
 [ln(2 5)ln(  )]
2 2 2 | M1
A1 | 1.1b | arsinh x=ln(x+(x2 + 1))
correct expression | (used)
1 2( 52)
 ln
2 51 | M1 | 2.1 | combining lns
1 2( 52)( 51)
 ln
2 ( 51)( 51) | M1 | 2.1 | rationalizing denominator
(must be seen)
1 ( 53)
 ln *
2 2 | 1 ( 53)
 ln *
2 2 | A1cao | 2.1 | NB AG
(5)
[8]
Alternative solution
x + ay = 2, x + ay + z = 1
 2x + z = 1, z = 2x 1
x + ay = 2  y = (2+ x)/a
3x6
 2x 12x3
a
2a6 2 4a9
x y , z = 
3 3 3
M1
M1
M1
A3
[6]
from 2 equations
to get eqn in one unknown
DR
3 3
1 1
2 dx2 dx
3 3
4x2 4x2 (2x1)21
4 4
M1
3.1a
3
1 1
or 2 dx
2 3 (x1/2)21/4
4
1.1b
1.1b
M1
A1
arsinh x=ln(x+(x2 + 1))
correct expression
rationalizing denominator
(must be seen)
15 In this question you must show detailed reasoning.
Show that $\int _ { \frac { 3 } { 4 } } ^ { \frac { 3 } { 2 } } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 4 x + 2 } } \mathrm {~d} x = \frac { 1 } { 2 } \ln \left( \frac { 3 + \sqrt { 5 } } { 2 } \right)$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q15 [8]}}
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